$f(x, y) = x^2 - y^2$ We have a change of variables: $\begin{aligned} x &= X_1(u, v) = \dfrac{u}{2} \\ \\ y &= X_2(u, v) = \dfrac{u}{2} - \dfrac{v}{2} \end{aligned}$ What is $f(x, y)$ under the change of variables? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{u^2}{2} + \dfrac{uv - v^2}{3}$ (Choice B) B $\dfrac{u^2}{2} + \dfrac{uv}{2} - \dfrac{v^2}{4}$ (Choice C) C $\dfrac{uv}{2} - \dfrac{v^2}{4}$ (Choice D) D $\dfrac{uv - v^2}{3}$
Solution: When applying a change of variables, we substitute the new definition for $x$ and $y$ into the original equation. The original equation: $f(x, y) = x^2 - y^2$ Let's substitute $X_1(u, v)$ for $x$ and $X_2(u, v)$ for $y$. $\begin{aligned} f(x, y) &= \left( \dfrac{u}{2} \right)^2 - \left( \dfrac{u}{2} - \dfrac{v}{2} \right)^2 \\ \\ &= \dfrac{u^2}{4} - \left( \dfrac{u^2}{4} - \dfrac{uv}{2} + \dfrac{v^2}{4} \right) \\ \\ &= \dfrac{uv}{2} - \dfrac{v^2}{4} \end{aligned}$ Therefore, under the change of variables, $f(x, y)$ becomes: $\dfrac{uv}{2} - \dfrac{v^2}{4}$